Prove $\sum_{n=1}^{\infty}(\sqrt{n^2+3}-\sqrt{n^2-1})=\infty$ I want to prove $$\sum_{n=1}^{\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=\infty$$
If $\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\neq0$, I can prove it.
In fact, however, $$\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=0$$
As a solution to finding a sum of series, I only know one that uses a partial sum or a progression of differences. How can I apply it to this proof?
I would appreciate it if you could tell me the solution.
 A: You have$$\sqrt{n^2+3}-\sqrt{n^2-1}=\frac4{\sqrt{n^2+3}+\sqrt{n^2-1}}.$$Note that the denominator of the previous expression behaves as $n+n(=2n)$ when $n$ is large, and that therefore the whole fraction behaves as $\frac2n$. In fact,\begin{align}\lim_{n\to\infty}\frac{\frac4{\sqrt{n^2+3}+\sqrt{n^2-1}}}{\frac1n}&=\lim_{n\to\infty}\frac{4n}{\sqrt{n^2+3}+\sqrt{n^2-1}}\\&=\lim_{n\to\infty}\frac4{\sqrt{1+\frac3{n^2}}+\sqrt{1-\frac1{n^2}}}\\&=\frac42\\&=2.\end{align}The statement that you want to prove follows from this. Since that limit is $2$, then there is some $N\in\Bbb N$ such that$$n\geqslant N\implies\sqrt{n^2+3}-\sqrt{n^2-1}\geqslant\frac1n.$$So,$$\sum_{n=N}^\infty\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=\infty,$$and therefore$$\sum_{n=1}^\infty\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=\infty.$$
A: $\sqrt{n^2+3}-\sqrt{n^2-1}=$
$=\dfrac{\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\left(\sqrt{n^2+3}+\sqrt{n^2-1}\right)}{\sqrt{n^2+3}+\sqrt{n^2-1}}=$
$=\dfrac{n^2+3-(n^2-1)}{\sqrt{n^2+3}+\sqrt{n^2-1}}=\dfrac4{\sqrt{n^2+3}+\sqrt{n^2-1}}>$
$>\dfrac4{\sqrt{n^2+3n^2}+\sqrt{n^2}}=\dfrac4{\sqrt{4n^2}+\sqrt{n^2}}=$
$=\dfrac4{3n}>\dfrac1n\;,\quad$ for all $\;n\in\mathbb{N}\;.$
Hence,
$\sqrt{n^2+3}-\sqrt{n^2-1}>\dfrac1n\;,\quad$ for all $\;n\in\mathbb{N}\;.$
Since $\;\displaystyle\sum_{n=1}^\infty\dfrac1n=+\infty\;,\;$ by using the comparison test, we get that $\;\displaystyle\sum_{n=1}^{\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=+\infty\;.$
A: Also:
\begin{align}
\sqrt{n^2+1} &= n\left(1+\frac{1}{n^2}\right)^{1/2} = n\left(1+\frac{1}{2n^2} + O(n^{-4})\right)
\\ &= n + \frac{1}{2n} + O(n^{-3})
\\
\sqrt{n^2-3} &= n - \frac{3}{2n} + O(n^{-3})\qquad\text{similarly}
\\
\sqrt{n^2+1} - \sqrt{n^2-3} &= \frac{2}{n} + O(n^{-3})
\end{align}
Thus the series diverges by comparison with $\sum \frac{1}{n}$.

Sorry, I did $+1$ and $-3$ instead of $+3$ and $-1$.
A: Just to give a slightly different approach, note that for $n\ge1$, we have
$$\sqrt{n^2+3}-\sqrt{n^2-1}\gt\sqrt{n^2+2+{1\over n^2}}-\sqrt{n^2}=\left(n+{1\over n}\right)-n={1\over n}$$
so
$$\sum_{n=1}^\infty\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\gt\sum_{n=1}^\infty{1\over n}=\infty$$
